7.3SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES

OBJECTIVES

1. Decide whether an ordered pair is a solution of a linear system.

2. Solve linear systems by graphing.3. Solve linear systems by substitution.4. Solve linear systems by addition.5. Identify systems that do not have exactly one

ordered-pair solution.6. Solve problems using systems of linear

equations.

SYSTEM OF LINEAR EQUATIONS IN 2 VARIABLES

System of Linear Equations: -- 2 linear equations E.g., 2x – 3y = -4

2x + y = 4 Solution to a System of Equations:

-- an ordered pair that satisfy both equations E.g., (1, 2), or, x = 1 and y = 2 is a solution to

the system Check

2x – 3y = -4 2x + y = 42(1) – 3(2) = -4 2(1) + 2 = 42 – 6 = -4 true 2 + 4 = 4 true

GRAPHICAL INTERPRETATION OF THE SOLUTION

2x – 3y = -42x + 4 = 3yy = (2/3)x + (4/3)

2x + y = 4y = -2x + 4

(1, 2)

EXAMPLE

Solve by graphing the system of equations.

x + 2y = 2 an x – 2y = 6

Find x-intercept and y-intercept for both equations. x + 2y = 2 x – 2y = 6 when x = 0, when x = 0,0 + 2y = 2 0 – 2y = 6y = 1 y = -3

when y = 0 when y = 0,x + 2(0) = 2 x – 2(0) = 6x = 2 x = 6 (0, 1) -- y-intercept (0, -3) -- y-intercept(2, 0) -- x-intercept (6, 0) -- x-intercept

EXAMPLE (CONT.)

• Lines intersect at (4, -1).• Check:

x + 2y = 24 + 2(-1) = 24 – 2 = 2 true

x – 2y = 64 – 2(-1) = 64 + 2 = 6 true

SOLVING EQUATION SYSTEM BY SUBSTITUTION Solve:

y = -x – 14x – 3y = 24

Solution

Substitute the expression for y in the first equation for y in the second equation.

4x – 3y = 24 y = -x - 14x – 3(-x – 1) = 24 y = -(3) - 14x + 3x + 3 = 24 y = -47x = 21 x = 3 Solution: (3, -4)

SOLVING A LINEAR SYSTEM BY SUBSTITUTION

Solvey = -x – 14x – 3y = 24

This gives us: 4x – 3(−x – 1) = 24.Solving for x, we get: x = 3

Substitute x value back in the first equation.y = -(3) – 1

This gives us: y = -4 Solution: (3, -4)

EXAMPLE Solve the linear system.

-4x + y = -112x – 3y = 3

Solution1. Using one equation, express y in terms of x.

-4x + y = -11y = 4x - 11

2. Substitute this in the second equation.2x – 3y = 32x – 3(4x – 11) = 3

3. Solve for x2x – 12x + 33 = 3-10x = -30x = 3

EXAMPLE (CONT.)

4. Substitute this value of x in the first equation and solve for y.y = 4x – 11y = 4(3) – 11y = 12 – 11y = 1

Solution: (3, 1)

YOUR TURN

Solve the linear systemy = 2x + 72x – y = -5

Solution2x – y = -52x – (2x + 7) = -52x – 2x + 7 = -57 = -5

What does this mean? Check the slopes of the 2 lines.

SOLVING A LINEAR SYSTEM BY ADDITION Solve:

3x + 2y = 489x – 8y = -24

1. The idea is to eliminate either the x column or the y column and add the two equations.4(3x + 2y) = 4(48)9x – 8y = -24

2. 12x + 8y = 192 9x – 8y = -24

3. 21x = 1684. x = 8

SOLVING A LINEAR SYSTEM BY ADDITION

Substitute this value of x in either equation and solve for y.3x + 2y = 483(8) + 2y = 4824 + 2y = 482y = 24y = 12

Solution: (8, 12)

SPECIAL CASES

Number of Solutions Graphically

One ordered-pair solution Two lines intersect at one point.

No solution Two lines are parallel.

Infinitely many solutions Two lines are the same line.

A SYSTEM WITH NO SOLUTION

Solve:4x + 6y = 126x + 9y = 12

Using the addition method,4x + 6y = 126x + 9y = 12

0 = 12 false.

There is no solution to the system.

Multiply by 3

Multiply by -2

A SYSTEM WITH INFINITELY MANY SOLUTIONS

Solve:y = 3x – 215x – 5y = 10

Using the substitution method,15x – 5y = 1015x – 5(3x – 2) = 1015x – 15x + 10 = 1010 = 10

This is true for any (x, y) pairs. Thus, there is an infinitely number of

solutions.

MODELING WITH SYSTEMS OF EQUATIONS

Suppose a company produces and sells x iGizmos.

Revenue function: R(x) = (price per unit sold)x Cost function: C(x) = fixed cost

+ (cost per unit produced)x

Break-even point: intersection of R(x) and C(x)

C(x) = fixed cost + (price per unit produced) x

R(x) = fixed cost + (price per unit sold) x

Break-even point

x (iGizmo units)

Dolla

rs

FINDING A BREAK-EVEN POINT

A company plans to manufacture electronic age wheelchairs. Fixed cost will be $500,000, and the production cost for each wheelchair is $400. The chairs will be sold at $600 apiece.

Write the Cost function C(x). Write the Revenue function R(x). Graph the functions. Determine the beak-even point.

FINDING A BREAK-EVEN POINT

Cost function: C(x) = 500,000 + 400x Revenue function: R(x) = 600x Graph

FINDING A BREAK-EVEN POINT Break-even Point

C(x) = 500,000 + 400xR(x) = 600x

or

y = 500,000 + 400xy = 600x

600x = 500,000 + 400x200x = 500,000x = 2500

BREAK-EVEN POINT

x = 2500

y = 600x y = 600(2500) = 1,500,000

Thus, Break-even point: (2500, 1,500,000)

I.e., $1,500,000 with 2500 units sold.

YOUR TURN The profit function

P(x) = R(x) – C(x) For the preceding case,

P(x) = 600x – (500,000 + 400x)P(x) = 200x – 500,000

1. Sketch the graph of the profit function.2. What is the y-intercept of the function? How

do you interpret that value?3. What is the slope of the function? How do

you interpret that value?4. What is x-intercept of the function? How do

you interpret that value?